Re: [julia-users] Re: Best data structure to remove elements one by one

2014-02-25 Thread David P. Sanders
El domingo, 23 de febrero de 2014 18:09:36 UTC-6, Jeff Bezanson escribió: This was just changed on master so that you'd do Set([i*2 for i in 1:5]). OK, that seems reasonable, thanks. On Sun, Feb 23, 2014 at 12:45 AM, David P. Sanders dpsa...@gmail.comjavascript: wrote: El

[julia-users] Re: Best data structure to remove elements one by one

2014-02-22 Thread David P. Sanders
El viernes, 21 de febrero de 2014 16:03:19 UTC-6, Steven G. Johnson escribió: On Friday, February 21, 2014 9:02:48 AM UTC-5, David P. Sanders wrote: OK, I think I have answered my own question: a Set is the good structure. And to create a set from an array I can do something like s =

[julia-users] Re: Best data structure to remove elements one by one

2014-02-21 Thread David P. Sanders
El viernes, 21 de febrero de 2014 07:49:27 UTC-6, David P. Sanders escribió: Hi, Suppose I have a collection of elements that I will create once at the beginining, visit in an unknown order and delete one by one when they are visited. What is the most efficient data structure for this

[julia-users] Re: Best data structure to remove elements one by one

2014-02-21 Thread Steven G. Johnson
On Friday, February 21, 2014 9:02:48 AM UTC-5, David P. Sanders wrote: OK, I think I have answered my own question: a Set is the good structure. And to create a set from an array I can do something like s = Set([3, 4, 5]...) or s = Set([i*2 for i in 1:5]...) which would be the